3.95 \(\int \cos ^4(c+d x) \sqrt {a+a \cos (c+d x)} \, dx\)

Optimal. Leaf size=158 \[ \frac {2 a \sin (c+d x) \cos ^4(c+d x)}{9 d \sqrt {a \cos (c+d x)+a}}+\frac {16 a \sin (c+d x) \cos ^3(c+d x)}{63 d \sqrt {a \cos (c+d x)+a}}+\frac {32 \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{105 a d}-\frac {64 \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{315 d}+\frac {32 a \sin (c+d x)}{45 d \sqrt {a \cos (c+d x)+a}} \]

[Out]

32/105*(a+a*cos(d*x+c))^(3/2)*sin(d*x+c)/a/d+32/45*a*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+16/63*a*cos(d*x+c)^3*
sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+2/9*a*cos(d*x+c)^4*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)-64/315*sin(d*x+c)*(
a+a*cos(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.24, antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2770, 2759, 2751, 2646} \[ \frac {2 a \sin (c+d x) \cos ^4(c+d x)}{9 d \sqrt {a \cos (c+d x)+a}}+\frac {16 a \sin (c+d x) \cos ^3(c+d x)}{63 d \sqrt {a \cos (c+d x)+a}}+\frac {32 \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{105 a d}-\frac {64 \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{315 d}+\frac {32 a \sin (c+d x)}{45 d \sqrt {a \cos (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4*Sqrt[a + a*Cos[c + d*x]],x]

[Out]

(32*a*Sin[c + d*x])/(45*d*Sqrt[a + a*Cos[c + d*x]]) + (16*a*Cos[c + d*x]^3*Sin[c + d*x])/(63*d*Sqrt[a + a*Cos[
c + d*x]]) + (2*a*Cos[c + d*x]^4*Sin[c + d*x])/(9*d*Sqrt[a + a*Cos[c + d*x]]) - (64*Sqrt[a + a*Cos[c + d*x]]*S
in[c + d*x])/(315*d) + (32*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(105*a*d)

Rule 2646

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(-2*b*Cos[c + d*x])/(d*Sqrt[a + b*Sin[c + d*
x]]), x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2751

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(d
*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(f*(m + 1)), x] + Dist[(a*d*m + b*c*(m + 1))/(b*(m + 1)), Int[(a + b*Sin
[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] &&  !LtQ[m,
-2^(-1)]

Rule 2759

Int[sin[(e_.) + (f_.)*(x_)]^2*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(Cos[e + f*x]*(a
+ b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*(m + 2)), Int[(a + b*Sin[e + f*x])^m*(b*(m + 1) - a*
Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)]

Rule 2770

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[(-2*b*Cos[e + f*x]*(c + d*Sin[e + f*x])^n)/(f*(2*n + 1)*Sqrt[a + b*Sin[e + f*x]]), x] + Dist[(2*n*(b*c + a*d)
)/(b*(2*n + 1)), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, f}
, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[n, 0] && IntegerQ[2*n]

Rubi steps

\begin {align*} \int \cos ^4(c+d x) \sqrt {a+a \cos (c+d x)} \, dx &=\frac {2 a \cos ^4(c+d x) \sin (c+d x)}{9 d \sqrt {a+a \cos (c+d x)}}+\frac {8}{9} \int \cos ^3(c+d x) \sqrt {a+a \cos (c+d x)} \, dx\\ &=\frac {16 a \cos ^3(c+d x) \sin (c+d x)}{63 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a \cos ^4(c+d x) \sin (c+d x)}{9 d \sqrt {a+a \cos (c+d x)}}+\frac {16}{21} \int \cos ^2(c+d x) \sqrt {a+a \cos (c+d x)} \, dx\\ &=\frac {16 a \cos ^3(c+d x) \sin (c+d x)}{63 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a \cos ^4(c+d x) \sin (c+d x)}{9 d \sqrt {a+a \cos (c+d x)}}+\frac {32 (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{105 a d}+\frac {32 \int \left (\frac {3 a}{2}-a \cos (c+d x)\right ) \sqrt {a+a \cos (c+d x)} \, dx}{105 a}\\ &=\frac {16 a \cos ^3(c+d x) \sin (c+d x)}{63 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a \cos ^4(c+d x) \sin (c+d x)}{9 d \sqrt {a+a \cos (c+d x)}}-\frac {64 \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{315 d}+\frac {32 (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{105 a d}+\frac {16}{45} \int \sqrt {a+a \cos (c+d x)} \, dx\\ &=\frac {32 a \sin (c+d x)}{45 d \sqrt {a+a \cos (c+d x)}}+\frac {16 a \cos ^3(c+d x) \sin (c+d x)}{63 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a \cos ^4(c+d x) \sin (c+d x)}{9 d \sqrt {a+a \cos (c+d x)}}-\frac {64 \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{315 d}+\frac {32 (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{105 a d}\\ \end {align*}

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Mathematica [A]  time = 0.28, size = 92, normalized size = 0.58 \[ \frac {\left (1890 \sin \left (\frac {1}{2} (c+d x)\right )+420 \sin \left (\frac {3}{2} (c+d x)\right )+252 \sin \left (\frac {5}{2} (c+d x)\right )+45 \sin \left (\frac {7}{2} (c+d x)\right )+35 \sin \left (\frac {9}{2} (c+d x)\right )\right ) \sec \left (\frac {1}{2} (c+d x)\right ) \sqrt {a (\cos (c+d x)+1)}}{2520 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4*Sqrt[a + a*Cos[c + d*x]],x]

[Out]

(Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*(1890*Sin[(c + d*x)/2] + 420*Sin[(3*(c + d*x))/2] + 252*Sin[(5*(c
+ d*x))/2] + 45*Sin[(7*(c + d*x))/2] + 35*Sin[(9*(c + d*x))/2]))/(2520*d)

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fricas [A]  time = 0.72, size = 72, normalized size = 0.46 \[ \frac {2 \, {\left (35 \, \cos \left (d x + c\right )^{4} + 40 \, \cos \left (d x + c\right )^{3} + 48 \, \cos \left (d x + c\right )^{2} + 64 \, \cos \left (d x + c\right ) + 128\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{315 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+a*cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

2/315*(35*cos(d*x + c)^4 + 40*cos(d*x + c)^3 + 48*cos(d*x + c)^2 + 64*cos(d*x + c) + 128)*sqrt(a*cos(d*x + c)
+ a)*sin(d*x + c)/(d*cos(d*x + c) + d)

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giac [A]  time = 0.65, size = 129, normalized size = 0.82 \[ \frac {1}{2520} \, \sqrt {2} \sqrt {a} {\left (\frac {35 \, \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right )}{d} + \frac {45 \, \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )}{d} + \frac {252 \, \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )}{d} + \frac {420 \, \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )}{d} + \frac {1890 \, \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{d}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+a*cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

1/2520*sqrt(2)*sqrt(a)*(35*sgn(cos(1/2*d*x + 1/2*c))*sin(9/2*d*x + 9/2*c)/d + 45*sgn(cos(1/2*d*x + 1/2*c))*sin
(7/2*d*x + 7/2*c)/d + 252*sgn(cos(1/2*d*x + 1/2*c))*sin(5/2*d*x + 5/2*c)/d + 420*sgn(cos(1/2*d*x + 1/2*c))*sin
(3/2*d*x + 3/2*c)/d + 1890*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)/d)

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maple [A]  time = 0.18, size = 97, normalized size = 0.61 \[ \frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (560 \left (\cos ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-800 \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+552 \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-104 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+107\right ) \sqrt {2}}{315 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*(a+a*cos(d*x+c))^(1/2),x)

[Out]

2/315*cos(1/2*d*x+1/2*c)*a*sin(1/2*d*x+1/2*c)*(560*cos(1/2*d*x+1/2*c)^8-800*cos(1/2*d*x+1/2*c)^6+552*cos(1/2*d
*x+1/2*c)^4-104*cos(1/2*d*x+1/2*c)^2+107)*2^(1/2)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d

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maxima [A]  time = 1.13, size = 79, normalized size = 0.50 \[ \frac {{\left (35 \, \sqrt {2} \sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ) + 45 \, \sqrt {2} \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 252 \, \sqrt {2} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 420 \, \sqrt {2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 1890 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a}}{2520 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+a*cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

1/2520*(35*sqrt(2)*sin(9/2*d*x + 9/2*c) + 45*sqrt(2)*sin(7/2*d*x + 7/2*c) + 252*sqrt(2)*sin(5/2*d*x + 5/2*c) +
420*sqrt(2)*sin(3/2*d*x + 3/2*c) + 1890*sqrt(2)*sin(1/2*d*x + 1/2*c))*sqrt(a)/d

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\cos \left (c+d\,x\right )}^4\,\sqrt {a+a\,\cos \left (c+d\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^4*(a + a*cos(c + d*x))^(1/2),x)

[Out]

int(cos(c + d*x)^4*(a + a*cos(c + d*x))^(1/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**4*(a+a*cos(d*x+c))**(1/2),x)

[Out]

Timed out

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